圆的切线方程公式证明

圆的切线方程公式证明 已知:圆的方程为:(x - a)? + (y - b)? = r?, 圆上一点P(x0, y0) 解:圆心C(a, b) 直线CP的斜率:k1 = (y0 - b) / (x0 - a) 因为直线CP与切线垂直, 所以切线的斜率:k2 = -1/k1 = - (x0 - a) / (y0 - b) 根据点斜式, 求得切线方程: y - y0 = k2 (x - x0) y - y0 = [- (x0 - a) / (y0 - b)] (x - x0) 整理得:(x - x0)(x0 - a) + (y - y0)(y0 - b) = 0 (注意:这式也是很好用的切线方程公式) 展开后: x0x - ax + ax0 + y0y - by + by0 - x0? - y0? = 0 ~ (1) 因为点P在圆上, 所以它的坐标满足方程: (x0 - a)? + (y0 - b)? = r? 化简: x0? - 2ax0 + a? + y1? - 2by0 + b? = r? 移项: - x0? - y0? = -2ax0 - 2by0 + a? + b? - r? ~ (2) 由(2)代入(1), 得: x0x - ax + ax0 + y0y - by + by0 + (-2ax0 - 2by0 + a? + b? - r?) = 0 化简, (x0x - ax - ax0 + a?) + (y0y - yb - by0 + b?) = r? 整理, (x0 - a)(x - a) + (y0 - b)(y - b) = r? 类似地, 对於圆的一般方程:x? + y? + Dx + Ey + F = 0, 过圆上的点的切线方程. 2. 已知:圆的方程为:x? + y? + Dx + Ey + F = 0, 圆上一点P(x0, y0) 解:圆心C( -D/2, -E/2 ) 直线CP的斜率:k1 = (y0 + E/2) / (x0 + D/2) 因为直线CP与切线垂直, 所以切线的斜率:k2 = -1/k1 = - (x0 + D/2) / (y0 + E/2) 根据点斜式, 求得切线方程: y - y0 = k2 (x - x0) y - y0 = [- (x0 + D/2) / (y0 + E/2)] (x - x0) 整理得:x0x + y0y + Dx/2 + Ey/2 - Dx0/2 - Ey0/2 -x0? - y0? = 0 ~ (3) 因为点P在圆上, 所以它的坐标满足方程: x0? + y0? + Dx0 + Ey0 + F = 0 移项: - x0? - y0? = Dx0 + Ey0 + F ~ (4) 由(4)代入(3), 得: x0x + y0y + Dx/2 + Ey/2 - Dx0/2 - Ey0/2 + Dx0 + Ey0 + F = 0 整理, x0x + y0y + D(x + x0)/2 + E(y + y0)/2 + F = 0 3a. 已知:圆的方程为:(x - a)? + (y - b)? = r? , 圆外一点P(x0, y0) 解: 圆心C(a, b), 设切点为M 则切线长PM = ? (CP? - MC?) (根据勾股定理) = ? [(x0 - a)? + (y0 - b)? - r?] (CP:两点间距离公式求得, MC:半径长) 类似地, 对於圆的一般方程:x? + y? + Dx + Ey + F = 0, 过圆外的点的切线长.... 3b. 已知:圆的方程为:x? + y? + Dx + Ey + F = 0 , 圆外一点P(x0, y0) 解: 圆心C( -D/2, -E/2 ), 设切点为M 则切线长PM = ? (CP? - MC?) (根据勾股定理) = ? [ (x0 + D/2)? + (y0 + E/2)? - ((?(D?+E?-4F))/2)? ] (半径:r=(?(D?+E?-4F)) / 2) = ? (x0? + y0? + Dx0 + Ey0 + F)